Page 188 - Linear Algebra Done Right

Square Roots
Square Roots
Recall that a square root of an operator T ∈L(V) is an operator 177
2
S ∈L(V) such that S = T. As an application of the main structure
theorem from the last section, in this section we will show that every
invertible operator on a complex vector space has a square root.
Every complex number has a square root, but not every operator on
a complex vector space has a square root. An example of an operator
3
on C that has no square root is given in Exercise 4 in this chapter.
The noninvertibility of that particular operator is no accident, as we
will soon see. We begin by showing that the identity plus a nilpotent
operator always has a square root.

8.30 Lemma: Suppose N ∈L(V) is nilpotent. Then I + N has a
square root.

√
Proof: Consider the Taylor series for the function 1 + x:

2
8.31 1 + x = 1 + a 1 x + a 2 x + ··· . Because a 1 = 1/2, this
formula shows that
We will not find an explicit formula for all the coefficients or worry 1 + x/2 is a good
about whether the infinite sum converges because we are using this estimate for √ 1 + x
equation only as motivation, not as a formal part of the proof. when x is small.
Because N is nilpotent, N m = 0 for some positive integer m. In 8.31,
suppose we replace x with N and 1 with I. Then the infinite sum on
j
the right side becomes a finite sum (because N = 0 for all j ≥ m). In
other words, we guess that there is a square root of I + N of the form
2
I + a 1 N + a 2 N +· · ·+ a m−1 N m−1 .
Having made this guess, we can try to choose a 1 ,a 2 ,...,a m−1 so that
the operator above has its square equal to I + N. Now
2
3
(I+a 1 N + a 2 N + a 3 N + ··· + a m−1 N m−1 2
)
2 2 3
= I + 2a 1 N + (2a 2 + a 1 )N + (2a 3 + 2a 1 a 2 )N +· · ·
+ (2a m−1 + terms involving a 1 ,...,a m−2 )N m−1 .
We want the right side of the equation above to equal I + N. Hence
choose a 1 so that 2a 1 = 1 (thus a 1 = 1/2). Next, choose a 2 so that
2
2a 2 + a 1 = 0 (thus a 2 =−1/8). Then choose a 3 so that the coefficient
3
of N on the right side of the equation above equals 0 (thus a 3 = 1/16).